Cor. The difference between any two sides of a triangle is less than the third side. ABC is a A. It is required to show that AB - AC < BC. ABAC + BC. From each of these unequals take AC, and AB AC < BC. (I—16, p. 59.) In the same manner it may be shown that AB <AC and that BC AC < AB. 61.-Exercises 1. Show that the sum of any three sides of a quadrilateral is greater than the fourth side. 2. The sum of the four sides of a quadrilateral is greater than the sum of its diagonals. 3. The sum of the diagonals of a quadrilateral is greater than the sum of either pair of opposite sides. 4. The sum of the st. lines joining any point, except the intersection of the diagonals, to the four vertices of a quadrilateral, is greater than the sum of the diagonals. 5. If any point within a ▲ be joined to the ends of a side of the ▲, the sum of the joining lines is less than the sum of the other two sides of the A. 6. If any point within a ▲ be joined to the three vertices of the A, the sum of the three joining lines is less than the perimeter of the ▲, but greater than half the perimeter. 7. The sum of any two sides of a ▲ is greater than twice the median drawn to the third side. 8. The median of a ▲ divides the vertical into parts, of which the greater is adjacent to the less side. 9. The perimeter of a ▲ is greater than the sum of the three medians. 10. A and B are two fixed points, and CD is a fixed st. line. Find the point P in CD, such that PA + PB is the least possible; (a) When A and B are on opposite sides of CD; (b) When A and B are on the same side of CD. 11. A and B are two fixed points, and CD is a fixed st. line. Find the point P in CD, such that the difference between PA and PB is the least possible; AE (a) When A and B are on the same side of CD; (b) When A and B are on opposite sides of CD. 12. Prove Theorem 16 by producing BA to E, making AC, and joining EC. = 13. Prove that the shortest line which can be drawn with its ends on the circumferences of two concentric circles, will, when produced, pass through the centre. 14. Prove the Corollary under Theorem 16, (a) by cutting off from AB a part AD = AC and joining DC; (b) by producing AC to E making AE = AB and joining BE. THEOREM 17 If two triangles have two sides of one respectively equal to two sides of the other but the contained angle in one greater than the contained angle in the other, the base of the triangle which has the greater angle is greater than the base of the other. Hypothesis.—ABC, DEF are two ▲s having AB = DE, AC = DF and < BAC> < EDF. To show that BC > EF. Construction.-Make DG = EDG = BAC and cut off AC, or DF. Join EG. Bisect FDG and let the bisector meet EG at H. Proof.— In As ABC, DEG, In As FDH, GDH, Join FH. In A EHF, EH + HF > EF. (I—16, p. 59.) THEOREM 18 (Converse of Theorem 17) If two triangles have two sides of one respectively equal to two sides of the other but the base of one greater than the base of the other, the triangle which has the greater base has the greater vertical angle. Да B Hypothesis.-ABC, DEF are two As having AB = DE, |